What if the quadratic expression in an inequality has no real zeroes? Suppose that you have to find the values of x which satisfy \({x^2} + x + 1 > 0\). The expression \({x^2} + x + 1\) has no real zeroes. If you plot its graph, you will obtain a parabola which lies entirely above the horizontal axis:

It is easy to see that \({x^2} + x + 1 > 0\) for all values ofx. Thus, the solution set of \({x^2} + x + 1 > 0\) is the set of all real numbers.

On the other hand, the solution set of \({x^2} + x + 1 < 0\) will be empty, because the value of \({x^2} + x + 1\) is never negative.

Solved Example 6: Let \(f\left( x \right)\) be a quadratic expressions. Suppose that the value of this expression is evaluated at three different points: \(x = a,\,\,x = b,\,\,x = c\), where \(a < b < c\). It is found that

Solution: This problem can be solved without any calculations. Now that since \(f\left( a \right) > 0\) and \(f\left( b \right) < 0\), the parabola of the expression must lie above the x-axis for \(x = a\) and below that x-axis for \(x = b\). This means the parabola must cross the x-axis between \(x = a\) and \(x = b\) (as indicated by the heavy point):

Similarly, \(f\left( b \right) < 0\) and \(f\left( c \right) > 0\), the parabola must cross the axis again between and :

Thus, not only can we say that \(f\left( x \right)\) has real and distinct zeroes, we can also specify the location of these zeroes: one zero is in the interval \(\left( {a,b} \right)\) and the second zero is in the interval \(\left( {b,c} \right)\). This graphical approach to problem solving in quadratics is extremely powerful!